AoPS Academy
such that 
for all 
such that for every 
is satisfied? Explain the answer.
be a scalene oblique triangle, and 
a point on the orthocentroidal circle of (
)., trilinear polar of the polar conjugate (
-isoconjugate) of , Droz-Farny axis of are concurrent. with respect to is as follows:
, there exists a pair of orthogonal lines 
, 
through such that the midpoints of the 3 segments cut off by , from the sidelines of are collinear. The line through these 3 midpoints is the Droz-Farny axis of wrt .
with nonnegative integer coefficients satisfies 
and 
. Find 
.
be a triangle, 
is incenter of triangle . Draw 
perpendicular to 
at 
and 
perpendicular to 
at 
, 
. Prove that: Area of triangle 
written on the board. We apply the following rule: At each step, we will add all the numbers that are the sum of the two numbers on the board so that the sum does not appear on the board. For example, if the two initial numbers are 
; then the numbers on the board after step 1 are 
; after step 2 are 
; 
, prove that the number 2024 cannot appear on the board.
; 
, prove that the number 2024 can appear on the board.
to 
on a different unit square of a 
square grid, such that consecutive integers are on adjacent squares, and is not adjacent to ? (Note that adjacent squares are squares that share a common side.)
be a prime and 
be positive integers such that 
and 
is prime. Prove that 
, 
, 
, and 
. The railway forms a complete loop, as shown below, and trains go in both directions. Suppose that a trip between two adjacent towns costs one ticket. Using exactly eight tickets, how many distinct ways are there of traveling from town and ending at town ? (You may pass through town in the middle).
. Then we get that: 
,so 
and hence 
, which is a contradiction by 
. Then we get that: 
With the same way as case 
we get that 
and hence 
, which is a contradiction by 
.
. Then we get that: 
, and hence 
, so equivalently we have that 
, which has only 
as a positive integer solution. So, in this case 
,which is the only solution of the given equation.
and 
.
.
.
but, 
which is contradiction.
. then, 
.
.
. obviously contradicition.
.
yields 
is even, so 
. Now if 
we have ![\[2x\geq \nu_{2}(2020^x+2^y)=\min\{2x, 2y_1\}=\nu_{2}(2024^z)=3z.\]](http://latex.artofproblemsolving.com/e/6/3/e631c0799f33d41fc92bc875e0e69e3f038c1ba8.png)
However, 
as 
. Therefore 
, and the equation becomes 
. Modulo 
implies 
is odd, at which point 
. Comparing the 
's of both sides gives 
, so 
, 
for some nonnegative integer 
. Clearly, 
leads to the solution 
. When 
we derive a contradiction from:![\[1>\frac{2020^{3t+1}}{2024^{2t+1}}=\frac{505}{506}\cdot \left(\frac{2020^3}{2024^2}\right)^t\geq \frac{505}{506}\cdot \left(\frac{2020^3}{2024^2}\right)>1.\]](http://latex.artofproblemsolving.com/c/b/2/cb24c7a9ef36eabb403e6fa4982babe2ac46b8b8.png)
, which obviously works. We now prove it is the only one.
.
to get 
. Clearly, we must have 
, so is even and 
is odd. Let 
, for a positive integer 
. Then, the given equation becomes 
. Note that 
is always odd. Since 
and 
are both even, it follows if 
, then either 
or 
, neither of which are odd. Thus, 
. Because 
, we must have 
, and our claim follows. is .
. Note that when , 
is divisible by all the prime factors of 
, which are 
, , and 
. If 
, by Zsigmondy, there exists some 
that divides 
, and hence it is impossible that 
. Thus, , and we are done. 
, which clearly works.
gives 
. Since 
, if divided either one of or , then we have that one of 
is 
, which is absurd. Hence 
.
are both not multiples of , so they cannot be equal to 
. If 
, then we would have 
, which is absurd, so 
, so 
. Now we have ![\[ 2020^x + 4^x = 2024^z \]](http://latex.artofproblemsolving.com/4/b/e/4be984ca8fd9e9a9078af51ca885348c8cc030f5.png)
If 
, then by Zsigmondy there exists a prime not dividing 
that divides 
, absurd. Hence 
must hold.
Working in modulo 
, we obtain:
is 
for all 
.
is 
is 
and 
.
, where 
have the same parity 
.
Working in modulo , we obtain:
is odd number 
is odd number.
.
by 
results:
.
is an even non-negative number and results: the last digit of 
is 
.
must be , hence 
and we obtain
.
in the last relation, results:
is solution of the equation 
.
. by 
results:
.
is odd number; 
and 
are even numbers, hence the last equation has no solutions.
..
, odd number (see 
).
, even number, hence the equation has no solutions.
, even number, hence the equation has no solutions.
.
, where 
(see ).
.
.
.
, contradiction since
(proved many times in the previous posts).
has the unique solution .
. Because of this, we know y must be even and z must be odd. and 
. Prime factorizing the original equation, 
is the smallest factor, then 
, impossible. Similarly, if 2x is the largest factor of 2, 
is impossible. This means 
and the factor of 2 on the LHS is 
since the v2 of 
is at 2. (since its 0 mod 2 and not 0 mod 4). So we have 
. since 
is an integer, set 
Plugging in and dividing by largest factor of 2 we get 
k=0 works, but any larger k dosent work since 
so the only solution is k=0, which represents 
even and 
odd
we get:
if even then: 
which is a contradiction so 
even we get that 
so our equation transforms into the following:
Claim: 
which is a contradiction because we found that 
Claim: 
on both sides and using Lifting the Exponent Lemma (LTE) we get:
Since 
we get that 
Let 
also 
Now by taking we get:
but this isn't true for 
so 
clearly 
so 
is the only solution 
.
is even
gives is odd
is same as 
nonsense
nonsense
again we get is odd
(for a non-negetive integer k)
for 
for 
, so is even and is odd. Now consider of both sides. The RHS is 
, so it's odd. The LHS is 
, and it's odd if and only if . Now write the LHS as 
, and taking mod 4 on the inside sum gives 
. Now we have the relationship 
. We can easily check that 
is a solution, and we see that for , the LHS is strictly greater than the RHS, so there are no solutions 
.
![\[
2020^x + 2^y = 2024^z
\]](http://latex.artofproblemsolving.com/1/e/5/1e5535fd71c16a98e46030aa8eae8404de4804b5.png)
![\[
v_2(2020^x + 2^y) = v_2(2024^z)
\]](http://latex.artofproblemsolving.com/4/0/3/403f86d0ea5bb9977b3bda19223adb3e9c62260f.png)
![\[
v_2(2020) = 2 \Rightarrow v_2(2020^x) = 2x, \quad v_2(2^y) = y, \quad v_2(2024) = 4 \Rightarrow v_2(2024^z) = 4z
\]](http://latex.artofproblemsolving.com/c/2/0/c20aac682f210f5a4fcc284430d5eee48e3bdf1f.png)
![\[
2020^x + 2^y = 2^y(a + 1), \quad \text{where } a \in \mathbb{N}, \text{ odd}
\]](http://latex.artofproblemsolving.com/0/f/2/0f2148ff5c9fa26d0a4fd71a250d11c8771c6e1d.png)
So,![\[
v_2(2020^x + 2^y) = y = 4z \Rightarrow 2x = 4z \Rightarrow x = 2z
\]](http://latex.artofproblemsolving.com/e/b/1/eb16f0372960c1aed9b43478173ad5131e365f96.png)
![\[
2^{2x} + 1 \equiv 0 \pmod{3} \Rightarrow 2^{2x} \equiv -1 \pmod{3}
\]](http://latex.artofproblemsolving.com/c/e/3/ce36c0bd7856b2d5c2c9807b7c408ccbdfc742ac.png)
![\[
2^{2x} \equiv 1 \pmod{3} \Rightarrow \text{Contradiction}
\]](http://latex.artofproblemsolving.com/d/1/7/d1704603ecc3666f2d0498c3f981936d10fd52c9.png)
![\[
2020^x + 2^y = 2^{2x}(a + 1), \quad \text{where } a \text{ is odd}
\Rightarrow v_2 = 2x
\Rightarrow 4z = 2x \Rightarrow z = \frac{x}{2}
\Rightarrow x \text{ must be even}
\]](http://latex.artofproblemsolving.com/c/e/f/cef5778173248f0b5ba255751ce2b51b2cf5bbe0.png)
![\[
2020^x + 2^{2x} = 2024^z
\Rightarrow 4^x(505^x + 1) = 2^{4z}(253^z)
\]](http://latex.artofproblemsolving.com/7/c/9/7c99831ed753bc3fee358d0b7b10c175d0fce540.png)
![\[
2^{2x}(505^x + 1) = 2^{4z}(253^z)
\Rightarrow 505^x + 1 = 2^{2z}(253^z)
\]](http://latex.artofproblemsolving.com/b/9/3/b93b102a69645ef2f1c7e9211a6bb9be85d1a752.png)
![\[
2020^1 + 2^2 = 2020 + 4 = 2024 = 2024^1 \Rightarrow \text{Valid solution}
\]](http://latex.artofproblemsolving.com/6/8/a/68a1da7eb19ad5b3baff0ef83acc0bb5907a636f.png)
![\[
x = 1,\quad y = 2x = 2,\quad 2x + 1 = 3z \Rightarrow z = \frac{2x + 1}{3} = \frac{3}{3} = 1
\]](http://latex.artofproblemsolving.com/7/6/9/769beb4f38b4409b15dd5e6f7a8373d3011f2812.png)
![\[
2020^2 + 2^4 = 4080400 + 16 = 4080416 \\
2024^2 = 4096576 \Rightarrow \text{LHS} < \text{RHS}
\]](http://latex.artofproblemsolving.com/8/a/a/8aa1b5c9ada5421a8ffcf3e48364666b6a29b81e.png)
![\[
2020^3 + 2^6 = 8204080000 + 64 = 8204080064 \\
2024^3 = 8283809024 \Rightarrow \text{LHS} < \text{RHS}
\]](http://latex.artofproblemsolving.com/f/3/6/f360a4f6f8552bf9d485a5f9c658131fa6ccb364.png)
![\[
\boxed{(x, y, z) = (1, 2, 1)}
\]](http://latex.artofproblemsolving.com/2/2/0/220c534c815104162b53ed786d25a370af5ccfdf.png)
.![\[P=\dfrac{a}{\sqrt{2bc+1}}+\dfrac{b}{\sqrt{2ca+1}}+\dfrac{c}{\sqrt{2ab+1}}.\]](http://latex.artofproblemsolving.com/1/f/3/1f39a54e4cf378bfd4e66d97949350b9b37dc1ec.png)
and 
where 
denotes greatest integer function.
and 
and 
be distinct real numbers such that 
and 
that satisfy the equation
we get:
![$2020^x=2024^z-8^z=2016[...]$](http://latex.artofproblemsolving.com/3/8/e/38eaa7c52b664c59165d40a2823dc5b2d0416c86.png)
contradiction since 
but not 
or 
we get that 
and then from inelyalites esily get 
,
and we claim there are no more solutions. Our strategy relies on that the powers of 
is the minimal among the three, then we have a contradiction via mod 
.
and 
.
.
,
.
(as 
for 
and 
for some positive integer 
.
,
,
.
.
is even, but not divisible by 
.
,
.
,
,
and we claim there are no more solutions. Have in mind that 
and 
.
:
and hence, by computations up to 
or similar, 
for some positive integer 
.
(we used 
by Fermat's Little Theorem) and hence, by computations up to 
or similar, 
:
,
.
.
now imply that 
and 
and later by 
)